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An entire function $f$ obeys the estimate $|f(z)|\leq me^{\alpha x}$ for all points $z=x+iy$, where $m$ and $\alpha$ are positive constants. Verify that $f$ has the form $f(z)=Ae^{\alpha z}$ for some constan A. Could one draw the same conclusion if one knew only that $|f(z)|\leq me^{\alpha |z|}$ held for every $z$?

I am trying to solve this problem using the following:

Property of Entire Functions

I know that $|me^{\alpha z}|=me^{\text{Re}(\alpha z)}=me^{\alpha x}$ and so $|f(z)|\leq |me^{\alpha z}|$ with which by the above post $f(z)=cme^{\alpha z}$, then $f(z)=Ae^{\alpha z}$ where $A:=cm$. Is this reasoning correct? Thank you very much.

How to answer the question they ask in the exercise?

user482152
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2 Answers2

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For the first one you could alternatively say: \begin{align} \bigg|\frac{f(z)}{ e^{\alpha z}}\bigg| \leq m \end{align} Since $e^{\alpha z}$ is nonzero. So we have $f(z) e^{-\alpha z}$ is enitre as well. So by Liouville we have: \begin{align} f(z)e^{-\alpha z} = A \end{align} Hence: \begin{align} f(z) = Ae^{\alpha z} \end{align} The second one is not true. Take $f(z)=1$ and you have: \begin{align} 1 \leq e^{|z|} \end{align} For $m=1$ and $\alpha=1$

Shashi
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Yes, your reasoning is correct. To give an explicit answer, condsider the entire function

$$ g(z) = \frac{f(z)}{me^{\alpha \cdot z}} $$

By hypothesis, $|g(z)| \leq 1$, and so applying Liouville's theorem, $g(z)$ must be constant. This answers the first question. As for the second one, $f = 1$ is a counterexample with $m = \alpha = 1$.

qualcuno
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